[latexrefman] \( (x_0+x_1)^2 \leq (x_0)^2+(x_1)^2 \)

Vincent Belaïche vincent.belaiche at gmail.com
Sat Aug 20 20:06:48 CEST 2022


Ooops:  you need x_0 & x_1 to have *OPPOSITE* sign for this to hold

Le sam. 20 août 2022 à 19:00, Vincent Belaïche
<vincent.belaiche at gmail.com> a écrit :
>
> Dear Jim,
>
> I am still propagating your r641, and I noted that you have changed \(
> (x_0+x_1)^2 \) to \( (x_0+x_1)^2 \leq (x_0)^2+(x_1)^2 \) in node
> Subscripts & superscripts.
>
> I feel uncomfortable with examplifying with an inequation that is not true
> for all x_0 & x_1 real (you need x_0 & x_1 to have the same sign for
> this to hold), a strange or controversial example distracts the reader.
>
> I propose either (in order to my descending preference):
>
> - to change it to \( (x_0+x_1)^2 = (x_0)^2+(x_1)^2 + 2 x_0 x_1 \)
> - to revert to \( (x_0+x_1)^2 \)
> - to change it to \( \left(\frac{x_0+x_1}{2}\right)^2} \leq
> \frac{(x_0)^2 + (x_1)^2}{2} \)
> - to change it to \( \sqrt{(x_0+x_1)^2} \leq \sqrt{(x_0)^2}+\sqrt{(x_1)^2} \)
>
>   V.



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